But if , then wouldn't it be ? Which would exceed the boundaries of ?[/tex]

No. We showed if then . This is a true statement, you can look at the above proof or draw the intervals on the number line. What you are saying is that if then but you need to remember that when it does can contain all the values of it contains those values of such that so it never gets out of its interval.

Wasn't ?

Sorry, I was not careful there.
You can write, where and to get your result now.

Boundaries

Ok, I was having the wrong concept of the triangle inequality. Thanks for clearing that up! Can't believe I overlooked it! Now I'm still confused about the boundaries of

Originally Posted by Plato

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Originally Posted by ThePerfectHacker

No. We showed if then . This is a true statement, you can look at the above proof or draw the intervals on the number line. What you are saying is that if then but you need to remember that when it does can contain all the values of it contains those values of such that so it never gets out of its interval.

Ok, I see that when it contains all the values from .

But aren't we trying to fulfill the inequality ? If then let's say that when , it will fulfill but not which was our original intention.

But aren't we trying to fulfill the inequality ? If then let's say that when , it will fulfill but not which was our original intention.

Similarly for , let's say , then it would fulfill but not .

Did I get my concepts wrong somewhere?

Ifthen (for ).

That is all we proved. You are saying but some values in do not satisfy . So what? You are thinking that ifthen. This is clearly not true, just take . When we restrain all the inequalities used in the proof becomes satisfied. But this does not mean that any other additional numbers which satisfy the other inequalities must satisfy our first inequality that we started with. We do not have to worry about that because we have constrained.